Circle A has a radius of 5  and a center of (8 ,2 ). Circle B has a radius of 3  and a center of (3 ,7 ). If circle B is translated by <4 , 8 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?

Jul 2, 2018

$\text{no overlap } \approx 5.01$

Explanation:

$\text{What we have to do here is compare the distance d }$
$\text{between the centres to the sum of the radii}$

• " if sum of radii">d" then circles overlap"

• " if sum of radii"< d" then no overlap"

$\text{Before calculating d we require to find the new centre}$
$\text{of B under the given translation}$

$\text{under the translation } < 4 , 8 >$

$\left(3 , 7\right) \to \left(3 + 4 , 7 + 8\right) \to \left(7 , 15\right) \leftarrow \textcolor{red}{\text{new centre of B}}$

$\text{to calculate d use the "color(blue)"distance formula}$

•color(white)(x)d=sqrt((x_2-x_1)^2+(y_2-y_1)^2)

$\text{let "(x_1,y_1)=(8,2)" and } \left({x}_{2} , {y}_{2}\right) = \left(7 , 15\right)$

$d = \sqrt{{\left(7 - 8\right)}^{2} + {\left(15 - 2\right)}^{2}} = \sqrt{1 + 169} = \sqrt{170} \approx 13.01$

$\text{sum of radii } = 5 + 3 = 8$

$\text{Since sum of radii"< d" then no overlap}$

$\text{min distance "=d-" sum of radii}$

$\textcolor{w h i t e}{\times \times \times \times \times} = 13.01 - 8 = 5.01$
graph{((x-8)^2+(y-2)^2-25)((x-7)^2+(y-15)^2-9)=0 [-40, 40, -20, 20]}